anonymous
  • anonymous
If y = 5/2(e^(x/5) + e^(-x/5)), prove that y'' = y/25.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I expanded the bracket: y = 5/2(e^(x/5)) + 5/2(e^(-x/5))
anonymous
  • anonymous
Then I took the first derivative:
anonymous
  • anonymous
y ' = 5x/10(e^(x/5)) - 5x/10(e^(-x/5))

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anonymous
  • anonymous
Now how should I go about with getting the second derivative? Should I factor out the 5x/10 and then use product rule?
anonymous
  • anonymous
\[y'=5/2[e ^{x/5}(1/5)+e^{-x/5}(-1/5)\]
anonymous
  • anonymous
\[y'=(5/2)(1/5)[e ^{x/5}-e ^{-x/5}]\]
anonymous
  • anonymous
Does it make a difference if we were to expand first with 5/2, and then take the first derivaitve?
anonymous
  • anonymous
not at all
anonymous
  • anonymous
Because that is what I did and then I took the dy/dx of each of the two terms.
anonymous
  • anonymous
but y did u multiply with x/5?
anonymous
  • anonymous
instead it is 1/5 the coefficient of x/5
anonymous
  • anonymous
No sorry I did the following: y = 5/2(e^{x/5}) + 5/2(e^{-x/5}) y '= 5/2(1/5)*(e^{x/5}) + 5/2(-1/5)(e^{-x/5})
anonymous
  • anonymous
correct
anonymous
  • anonymous
So then: y = 1/2(e^{x/5} + e^{-x/5})
anonymous
  • anonymous
y' u mean?
anonymous
  • anonymous
yes sorry, y'
anonymous
  • anonymous
y' = 1/2(e^{x/5} + e^{-x/5})
anonymous
  • anonymous
n there must b negative between to terms
anonymous
  • anonymous
coz 1/5 is the common factor
anonymous
  • anonymous
y' = 1/2(e^{x/5} - e^{-x/5}), right?
anonymous
  • anonymous
yes:)
anonymous
  • anonymous
So then for y'' I get..
anonymous
  • anonymous
\[y''=1/2[e ^{^{x/5}}(1/5)-e ^{-x/5}(-1/5)]\]
anonymous
  • anonymous
Yes I got that, then I expanded, and I got: y'' = 1/10(e^{x/5} + e^{-x/5})
anonymous
  • anonymous
\[y''=1/10[e ^{x/5}+e ^{-x/5}]\]
anonymous
  • anonymous
again no need to expand
anonymous
  • anonymous
\[e ^{x/5}+ e ^{-x/5}=2y/5\]
anonymous
  • anonymous
y''=y/25
anonymous
  • anonymous
How did you get e^x/5 + e^-x/5 = 2y/5?
anonymous
  • anonymous
OHHH
anonymous
  • anonymous
I think I know
anonymous
  • anonymous
got it?
anonymous
  • anonymous
I got it thank you so much. So when you are taking derivative of a function that has a constant which is multipled by a bracket that can be differentiated, you just take the derivative of the terms inside the bracket and multiply by the constant in front of it?
anonymous
  • anonymous
right:)
anonymous
  • anonymous
Ok awesome. Thank you for your help.
anonymous
  • anonymous
my pleasure

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